Lecture_truss [Compatibility Mode]

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9/18/2013

1

Truss

Engineering Mechanics- Truss

Important Features of a Truss

• A truss consists of long and slender straight

members connected at joints. No member is

continuous through a joint.

• Bolted or welded connections (joints) are assumed

to be pinned together. Forces acting at the member ends reduce to a single force and no couple. Only

two-force members are considered.

• When forces tend to pull the member apart, it is in

tension. When the forces tend to compress the

member, it is in compression.


Formation of a truss is nothing more than

assemblage of triangles!



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External Loads on a TrussMembers of a truss are slender and not capable of supporting large lateral loads.

Loads must be applied at the joints. Members are light in weight and economical.


• Most structures are made of several trusses joined together to form a space

framework. Each truss carries those loads which act in its plane and may

be treated as a two-dimensional structure.

Various Types of Truss




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Simple Truss

• A simple truss is constructed by

successively adding two members and

one connection to the basic triangular

truss.

• A simple truss is not necessarily

made only of triangles. Althoughtriangles are always maintained in

practice


• In a simple truss, m = 2n - 3 where

m is the total number of members

and n is the number of joints.

basic truss

• A simple truss is always internally

rigid. Will not deform and collapse

under the action of load

Analysis of Trusses by the Method of Joints

• Dismember the truss and create a freebody

diagram for each member and pin.

• The two forces exerted on each member are

equal, have the same line of action, and

opposite sense.

• Forces exerted by a member on the pins or

joints at its ends are directed along the member

and equal and opposite.

• For simple Truss: m = 2n – 3Total unknowns

member forces = m = 5

reactions = r = 3

Total no. of equilibrium equations

2n = 8 (2 force equilibrium euations per joint)

m+r=2n

The truss is said to be statically determinate

and completely constrained.






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Problem -1

For the given loading , determine the zero force member in the truss shown

Engineers Mechanics- Truss

Problem 2

Using method of joints, determine the forces in the members of the truss shown

below




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Problem 2


FAB=0=FBC

FAD= -5 (C)

FBD= FBF

2 FBD (1.6/3.4) + 20 + FBE = 0

FBD = FBF = -34 (C)

B

FEB=12 (T)

FED= FEF

D

FDE + FDB (3/3.4) = 0

FDE = 30 (T) = FEF

Check

-FAD - FBD (1.6/3.4) =21

(reaction)

Problem 3

Using the method of joints,

determine the force in each

member of the truss.

SOLUTION:

• Based on a free-body diagram of the

entire truss, solve the 3 equilibrium

equations for the reactions at E and C.

• Joint A is subjected to only two unknown

member forces. Determine these from the

joint equilibrium requirements.

• In succession, determine unknown

member forces at joints D, B, and E from


• All member forces and support reactions

are known at joint C. However, the joint

equilibrium requirements may be applied

to check the results.




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Problem 3SOLUTION:

• Based on a free-body diagram of the entire

truss, solve the 3 equilibrium equations for the

reactions at E and C.

m6m12 N1000m24 N2000

0

E

M C

N000,10 E

x x C F 0 0 xC

y y C F N10,000 N1000- N20000

N7000 yC


Problem -3

• Joint A is subjected to only two unknown

member forces. Determine these from the


534

N2000 AD AB F F

C F

T F

AD

AB

N2500

N1500

• There are now only two unknown member

forces at joint D.

DA DE

DA DB

F F

F F

532

C F

T F

DE

DB

N3000

N2500




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• There are now only two unknown member

forces at joint B. Assume both are in

tension.

N3750

25001000054

54

BE

BE y

F

F F

C F BE N3750

N5250

375025001500053

53

BC

BC x

F

F F

T F BC N5250

• There is one unknown member force at joint E. Assume the member is in

tension.

N8750

37503000053

53

EC

EC x

F

F F C F EC N8750


Problem -3

• All member forces and support reactions are

known at joint C. However, the joint

equilibrium requirements may be applied to

check the results.

checks 087507000

checks 087505250

54

53

y

x

F

F

Problem -3




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Analysis of Trusses by the Method of Sections

• When the force in only one member or the

forces in a very few members are desired, the

method of sections works well.

• To determine the force in member BD, pass a

section through the truss as shown and create

a free body diagram for the left side.

• With only three members cut by the section,

the equations for static equilibrium may be

applied to determine the unknown member forces, including F BD.


Problem 4

A Fink roof truss is loaded as shown. Use method of section to determine the

force in members CD, CE and BD




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Problem 4


Problem 5

Use method of section to determine the force in members GJ and IK of the

truss shown


100kN

100kN

100kN

Inverted K truss



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11

Problem-5


100kN

100kN

100kN

G

Solution JX = -300

KY = -JY = 100*(2.7 + 5.4 +8.1)/7.5 = 1620/7.5 = 216

Section XX, lower FBD,

∑MG = 0

-FKI [ 5.9( ] + JY(0.8) –JX(2.7) – KY(6.7) = 0

FKI = -143.19 kN (C )

∑MI = 0

FJG[ 5.9 ( ] + JY(6.7) –JX(2.7) – KY(0.8) = 0

FJG = 143.19 kN (T )

Compound Truss

• Compound trusses are composed of two or more

simple trusses . The trusses shown are statically

determinant, rigid, and completely constrained,

nr m 2




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Compound Truss

Determine the forces in bars 1. 2. and 3 of the plane truss supported and

loaded as shown in the figure

Problem 6


SOLUTION:

• Two simple trusses

ADE & BCF

• Detach to expose 6

unknowns• Draw free body

diagrams to find

forces in 1, 2 &3

E F

3



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Problem 6


Partially, Improperly and Properly Constrained Truss


• For a truss to be properly constrained:

– It should be able to stay in equilibrium for any combinationof loading.

– Equilibrium implies both global equilibrium and internalequilibrium.

• Note that if m + r < 2n , the truss is most definitely partially constrained (and is unstable to certainloadings). But m + r ≥ 2n, is no guarantee that thetruss is stable, and may be termed improperlyconstrained.

• If m + r < 2n, the truss can never be staticallydeterminate.



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• Fewer unknowns than

equations, partially

constrained

• Equal number unknowns

and equations but

improperly constrained


Inadequacy

of external

constraints

unstable

Simple Truss

(internally

rigid)




Check if partially constrained, fully constrained or improperly constrained

P

P

P

Statically determinate, properly constrained

A Simple Truss is always a rigid truss

m + r = 2n m=9, r = 3, n=6

m=9, r = 3, n=6m + r = 2n

improperly constrained (non rigid,

may collapse)

m + r < 2n

Partially constrained (non rigid, may

collapse)

m=8, r = 3, n=6



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Problem 7 : Partially, Improperly and Properly Constrained Truss

`

2n = 20, m = 16, r = 4.

The circled part is a simple truss that

is adequately supported with 3

reactions. So it is completely

constrained

We only have to worry about the

remaining portion. We can easily

show that no matter whatever loading

is applied at the joints of the

remaining part can be supported.

Completely constrained & Statically

determinate

Problem 7 : Partially, Improperly and Properly Constrained Truss


a) j = 7, m = 10, r = 3, 2j > m + r.

Partially Constrained

b) j = 8, r = 3, m = 13, 2j = m + r. Clearly

the truss cannot be equilibrium with thisloading.

Take equilibrium of joint C. BC is

tensile

Take a section as shown and moment

about left hinge is not balanced

Improperly Constrained.

BC

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